Question

If $A=\left[\begin{array}{cc}4& 3\\ 2& 5\end{array}\right]$, find $x$ and $y$ such that $A^2-xA+yI=0$

Answer 1

$A=\left[\begin{array}{cc}4& 3\\ 2& 5\end{array}\right]A^2=\left[\begin{array}{cc}4& 3\\ 2& 5\end{array}\right]\times \left[\begin{array}{cc}4& 3\\ 2& 5\end{array}\right]=\left[\begin{array}{cc}20& 27\\ 18& 31\end{array}\right]xA=\left[\begin{array}{cc}4x& 3x\\ 2x& 5x\end{array}\right]yI=\left[\begin{array}{cc}y& 0\\ 0& y\end{array}\right]A^2-xA+yI=0$

or $\left[\begin{array}{cc}20& 27\\ 18& 31\end{array}\right]-\left[\begin{array}{cc}4x& 3x\\ 2x& 5x\end{array}\right]+\left[\begin{array}{cc}y& 0\\ 0& y\end{array}\right]=0$

or $\left[\begin{array}{cc}20-4x+y& 27-3x\\ 18-2x& 31-5x+y\end{array}\right]=0$

or $27-3x=0$

or $x=9$

$20-4x+y=0$

or $20-36+y=0$

or $y=16$

$\therefore x=9,y=16$